anonymous

4 years ago

On average, when a basketball is dropped, it bounces 7/10 as high as it did on the previous bounce. If a ball is dropped from a height of 10 m, determine the total distance up and down that the ball travels by top of the 11th bounce. Round to the nearest hundredth

Are you familiar with the sum of a geometric series?
Let
\[S = 1 + x + x^2 + x^3 +...+x^n\]
then
\[xS = x + x^2 + x^3 + ...+x^{n+1}\]
and so
\[S - xS = S(1-x) = 1 - x^{n+1}\]
So
\[S = \frac{1-x^{n+1}}{1-x}\]
That's the sum of a geometric series. Now examining your problem, we have to add the following:
\[ 10 + \frac{7}{10} \cdot 10 + \frac{7}{10}\cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + \left(\frac{7}{10}\right)^2 \cdot 10 + ... \]
so on and so forth. Noting that the distance traveled (one-way, either up or down) after the nth bounce is
\[\left(\frac{7}{10}\right)^n \cdot 10\]
and that everything except the first and last bounce is repeated, we can say that our total distance traveled is
\[\left[2\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n\cdot 10\right]- 10 - \left(\frac{7}{10}\right)^{11}\cdot 10\]
but this looks just like a geometric series with x = 7/10, so that means
\[\sum_{n=0}^{11} \left(\frac{7}{10}\right)^n = \frac{1-\left(\frac{7}{10}\right)^{12}}{1-\frac{7}{10}}\]
And the rest is just calculation... :)